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2020-03-06 取URL参数 #1
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function getUrlParams(url){
let paramsArray = url.split('?')[1].split('&');
let params = {};
for(let i=0;i<paramsArray.length;i++){
let param = paramsArray[i].split('=');
if(param.length === 2){
let key = param[0];
let value = param[1];
params[key] = value;
}
else{
let key = param[0];
params[key] = true;
}
}
return params;
}
let params = getUrlParams('http://domain.com?page=10&action=list&lazy');
console.log(params.page);
console.log(params.action);
console.log(params.lazy); |
function getUrlParams(url) { |
还有很多场景没处理到,像是 url 本身不合法,parseInt 没办法处理精确数字不合法等
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`var url = 'http://www.inode.club?page=10&action=list&node';
} const newobj = getUrlParams(url); |
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function getUrlParams(url) {
if (typeof url !== 'string') {
throw new Error('the type of `url` must be a string.');
}
if (url.indexOf('?') > 0) {
query = url.substr(url.indexOf('?') + 1)
}
if (!query) return {};
var params = {};
var items = query.split('&');
var len = items.length;
while (len-- > 0) {
var _name = items[len].split('=')[0];
var value = items[len].split('=')[1] || true;
params[_name] = value;
}
return params;
}
var url = 'http://domain.com?page=10&action=list&lazy';
var params = getUrlParams(url)
console.log(JSON.stringify(params, null, 4)) |
今天第一天先来一道简单的
在日常开发中,我们经常会遇到取url上面参数的需求,今天这道题就是取URL上面的参数。
不需要考虑浏览器兼容性
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